Answer:
a
[tex]A = 0.081 \ m[/tex]
b
The value is [tex]u = 0.2569 \ m/s[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.750 \ kg[/tex]
The spring constant is [tex]k = 17.5 \ N/m[/tex]
The instantaneous speed is [tex]v = 39.0 \ cm/s= 0.39 \ m/s[/tex]
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k * A^2[/tex]
Here a is the amplitude of the subsequent oscillations
=> [tex]A = \sqrt{\frac{m * v^ 2 }{ k} }[/tex]
=> [tex]A = \sqrt{\frac{0.750 * 0.39 ^ 2 }{17.5} }[/tex]
=> [tex]A = 0.081 \ m[/tex]
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k x^2 + \frac{1}{2} * m * u^2[/tex]
=> [tex]\frac{1}{2} * 0.750 * 0.39^2 = \frac{1}{2} * 17.5* 0.750(0.081 )^2 + \frac{1}{2} * 0.750 * u^2[/tex]
=> [tex]u = 0.2569 \ m/s[/tex]
